Difference between revisions of "1959 AHSME Problems/Problem 45"
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== Solution == | == Solution == | ||
| − | <math>\boxed{\textbf{(B)}}</math> | + | From the properties of [[logarithms]], we can simplify the equation and solve for <math>y</math>: |
| + | \begin{align*} | ||
| + | (\log_3 x)(\log_x 2x)(\log_{2x} y) &= \log_{x}x^2 \\ | ||
| + | (\log_3 2x)(\log_{2x} y) &= 2\log_x x \\ | ||
| + | \log_3 y &= 2 \\ | ||
| + | y &= 3^2 \\ | ||
| + | y &= 9 | ||
| + | \end{align*} | ||
| + | Thus, our answer is <math>\boxed{\textbf{(B) }9}</math>. | ||
== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=44|num-a=46}} | {{AHSME 50p box|year=1959|num-b=44|num-a=46}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 11:41, 22 July 2024
Problem
If 
, then 
equals:
Solution
From the properties of logarithms, we can simplify the equation and solve for :
\begin{align*}
(\log_3 x)(\log_x 2x)(\log_{2x} y) &= \log_{x}x^2 \\
(\log_3 2x)(\log_{2x} y) &= 2\log_x x \\
\log_3 y &= 2 \\
y &= 3^2 \\
y &= 9
\end{align*}
Thus, our answer is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 44 |
Followed by Problem 46 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
